\newproblem{lay:2_2_21}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.2.21}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Explain why the columns of an $n\times n$ matrix $A$ are linearly independent when $A$ is invertible.
}{
  % Solution
	If $A$ is invertible we have shown (see Theorem 2.2, Chapter 3, Biomedical Engineering Notes) that for every $\mathbf{b}\in\mathbb{R}^n$, there is
	a unique solution of the equation $A\mathbf{x}=\mathbf{b}$. In particular, there exists a solution for the equation $A\mathbf{x}=\mathbf{0}$ that is
	$\mathbf{x}=A^{-1}\mathbf{0}=\mathbf{0}$. Since the only solution of this problem is the trivial one, then by Theorem 6.1, Chapter 2, Biomedical Engineering Notes, 
	the columns of $A$ are linearly independent.
}
\useproblem{lay:2_2_21}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
